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ECE 461 Fall 2006

September 20, 2006Digital Modulation

After possible source and error control encoding, we have a sequence {mn} of message symbolsto be transmitted on the channel. The message symbols are assumed to come from a finitealphabet, say {0, 1, . . . ,M 1}. In the simplest case of binary signaling, M = 2. Each symbolin the sequence is assigned to one of M waveforms {s0(t), . . . , sM1(t)}.

Memoryless modulation versus modulation with memory. If the symbol to waveform mapping isfixed from one interval to the next, i.e., m 7 sm(t), then the modulation is memoryless. If themapping from symbol to waveform in the n-th symbol interval depends on previously transmittedsymbols (or waveforms) then the modulation is said to have memory.

For memoryless modulation, to send the sequence {mn} of symbols at the rate of 1/Ts symbolsper second, we transmit the signal

s(t) =n

smn(t nTs) . (1)

Linear versus nonlinear modulation. A digital modulation scheme is said to be linear if wecan write the mapping from the sequence of symbols {mn} to the transmitted signal s(t) asconcatenation of a mapping from the sequence {mn} to a complex sequence {cn}, followed by alinear mapping from {cn} to s(t). Otherwise the modulation is nonlinear.

Linear Memoryless Modulation

In this case, the mapping from symbols to waveforms can be written in complex baseband as:

sm(t) =Em ejmg(t) , m = 1, 2, . . . ,M , (2)

where g(t) is a real-valued, unit energy, pulse shaping waveform.

The signal sm(t) can be represented by a point in the complex plane, i.e., the signal spacecorresponding to a symbol interval is a 1-d (complex) space with basis function g(t).

PSfrag replacements

II

QQ

sm

Emm

s1

s2s3

s4sM

Representation of sm(t) Signal constellation

cV.V. Veeravalli, 2006 1

In real passband,sm(t) = Re[

2sm(t) e

j2pifct] =

2Em cos(2pifct + m)g(t) . (3)

As we saw in class, the signal energy is the same in both the real passband and complex basebanddomains and equals Em.

The average symbol energy for the constellation is given by

Es = 1M

Mm=1

Em . (4)

The average bit energy for the constellation (assuming that M = 2 , for some integer ) is givenby

Eb = Eslog2 M

=Es

. (5)

The distance between signals sk and sm is dk,m = sk sm, and the minimum distance is givenby

dmin = mink,m

dk,m . (6)

A measure of goodness of the constellation is the ratio

=d2minEb . (7)

Note that is independent of scaling of the constellation.

Some commonly used signal constellations are:

Pulse Amplitude Modulation (PAM). Information only in amplitude:

m = 0 andEm = (2m 1M)d

2, m = 1, 2, . . . ,M . (8)

We can compute as a function of M . For example, = 4 for M = 2.

Phase Modulation or Phase Shift Keying (PSK). Information only in phase:

m =2pim

Mand Em = E , m = 1, 2, . . . ,M . (9)

We showed in class that

dmin =

2E (1 cos 2piM)

= = 2 log2 M(1 cos 2piM

)

For QPSK, = 4 (as in BPSK).

Quadrature Amplitude Modulation (QAM). Information in phase and amplitude. We candesign constellations to maximize for a given M . Rectangular constellations are convenientfor demodulation. For rectangular 16-QAM, = 1.6.


Orthogonal Memoryless Modulation

Here the signal set is given by

sm(t) =E gm(t), m = 1, 2, . . . ,M (10)

where {gm(t)} are (possibly complex) unit energy signals, i.e., gm(t) = 1.

The correlation between signals sk(t) and sm(t) is given by:

km =sk(t), sm(t)

E = gk(t), gm(t) (11)

There are two kinds of orthogonality:

Orthogonality only in the real component of the correlation, i.e. Re{km} = 0, for k 6= m.This form of orthogonality is enough for coherent demodulation.

Complete orthogonality, i.e., km = 0, for k 6= m. This is required for noncoherent demodula-tion.

Examples of orthogonal signal sets Separation in time:

gm(t) = g (t (m 1)Ts/M) (12)where g(t) is such that g(t kTs/M), g(tmTs/M) = km. For example, g(t) = pTs/M (t), arectangular pulse of width Ts/M .

This signal set is completely orthogonal. We can also create a signal set of twice the size whichsatisfies orthogonality only in the real component of the correlation by adding {jgm(t)} to theabove signal set as we saw in class.

Separation in frequency:gm(t) = e

j2pi(m1)f t pTs(t) (13)

It is easy to show thatkm = sinc[Ts(k m)f ] ejpiTs(km)f (14)

and thatRe{km} = sinc[2Ts(k m)f ] . (15)

Thus the smallest value of f such that km = 0, for k 6= m, is 1/Ts, and such that Re{km} =0, for k 6= m, is 1/2Ts.

Separation in time and frequency: One way to do this is to pick {gm(t)} to be the Walsh-Hadamard functions on [0, Ts] as we saw in class.


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